1 Tính
a 204 - 84 : 12
b 5 mũ 6 : 5 mũ 3 + 2 mũ 3 x 2 mũ 2
c 15 x 2 mũ 3 + 4 x 3 mũ 2 - 5 x 7
d 164 x 53 + 47 x 164
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a) \(2^{4x}-3^{2x}=144-164:41\)
\(\Leftrightarrow16^x-9^x=144-4\)
\(\Leftrightarrow\) ......
bjan tự tính tiếp đc òi
\(A=5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9\)
\(A=5\cdot\left(2^2\right)^{15}\cdot\left(3^2\right)^9-2^2\cdot3^{20}\cdot\left(2^3\right)^9\)
\(A=5\cdot2^{30}\cdot3^{18}-2^2\cdot3^{20}\cdot2^{27}\)
\(A=5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}\)
\(A=2^{29}\cdot3^{18}\cdot\left(5\cdot2^1\cdot1-1\cdot3^2\right)\)
\(A=2^{29}\cdot3^{18}\cdot\left(5-9\right)\)
\(A=-2^2\cdot2^{29}\cdot3^{18}\)
\(A=-2^{31}\cdot3^{18}\)
_______________
\(B=5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6\)
\(B=5\cdot2^9\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot\left(3^3\right)^6\)
\(B=5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}\)
\(B=2^{28}\cdot3^{18}\cdot\left(5\cdot1\cdot3-7\cdot2\cdot1\right)\)
\(B=2^{28}\cdot3^{18}\cdot\left(15-14\right)\)
\(B=2^{28}\cdot3^{18}\)
Ta có: \(A:B\)
\(=\left(-2^{31}\cdot3^{18}\right):\left(2^{28}\cdot3^{18}\right)\)
\(=\left(-2^{31}:2^{28}\right)\cdot\left(3^{18}:3^{18}\right)\)
\(=-2^3\cdot1\)
\(=-8\)
1. \(6x^3-8=40\\ 6x^3=48\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2
2. \(4x^5+15=47\\ 4x^5=32\\ x^5=8\\ \Rightarrow x\in\varnothing\left(\text{vì }x\in N\right)\)Vậy x ∈ ∅
3. \(2x^3-4=12\\ 2x^3=16\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2
4. \(5x^3-5=0\\ 5x^3=5\\ x^3=1\\ \Rightarrow x=1\)Vậy x = 1
5. \(\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)Vậy \(x\in\left\{5;6\right\}\)
6. \(\left(3x-2\right)^{20}=\left(3x-1\right)^{20}\\ \Rightarrow3x-2=3x-1\\ 3x-3x=2-1\\ 0=1\left(\text{vô lí}\right)\)Vậy x ∈ ∅
7. \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\\ \left(3x-1\right)^{10}=\left[\left(3x-1\right)^2\right]^{10}\\ \Rightarrow\left(3x-1\right)^2=3x-1\\ \left(3x-1\right)^2-\left(3x-1\right)=0\\ \left(3x-1\right)\left[\left(3x-1\right)-1\right]=0\\ \left(3x-1\right)\left(3x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-1=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=1\\3x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\left(\text{loại vì }x\in N\right)\\x=\frac{2}{3}\left(\text{loại vì }x\in N\right)\end{matrix}\right.\)Vậy x ∈ ∅
8. \(\left(2x-1\right)^{50}=2x-1\\ \left(2x-1\right)^{50}-\left(2x-1\right)=0\\ \left(2x-1\right)\left[\left(2x-1\right)^{49}-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^{49}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=1\\2x-1=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\left(\text{loại vì }x\in N\right)\\x=1\left(t/m\right)\end{matrix}\right.\)Vậy x = 1
9. \(\left(\frac{x}{3}-5\right)^{2000}=\left(\frac{x}{3}-5\right)^{2008}\\ \left(\frac{x}{3}-5\right)^{2008}-\left(\frac{x}{3}-5\right)^{2000}=0\\ \left(\frac{x}{3}-5\right)^{2000}\left[\left(\frac{x}{3}-5\right)^8-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(\frac{x}{3}-5\right)^{2000}=0\\\left(\frac{x}{3}-5\right)^8=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}-5=0\\\frac{x}{3}-5=1\\\frac{x}{3}-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}=5\\\frac{x}{3}=6\\\frac{x}{3}=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\cdot3=15\\x=6\cdot3=18\\x=4\cdot3=12\end{matrix}\right.\)Vậy \(x\in\left\{15;18;12\right\}\)
\(1.6x^3-8=40\\ \Leftrightarrow6x^3=48\\ \Leftrightarrow x^3=8\Leftrightarrow x^3=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{2;-2\right\}\)
\(2.4x^3+15=47\) (T nghĩ đề là mũ 3)
\(\Leftrightarrow4x^3=32\Leftrightarrow x^3=8=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{2;-2\right\}\)
Câu 3, 4 tương tự nhé.
1,a) 695- [200+ (11- 12)]
= 695- [200+ (11- 1)]
= 695- [200+ 10]
= 695- 210
= 485
b) (519: 517+ 3): 7
= (52+ 3): 7
= (25+ 3): 7
= 28: 7
= 4
c) 129- 5[29- (6- 12)]
= 129- 5[29- (6- 1)]
= 129- 5[29- 5]
= 129- 5. 24
= 129- 120
= 9
3,a) 2x- 49= 5. 32
2x- 49= 5. 9
2x- 49= 45
2x = 45+ 49
2x = 94
x = 94: 2
x = 47
c) 2x- 15= 17
2x = 17+ 15
2x = 32
2x = 25
=> x = 5
Câu 3b bạn tự làm nhé, xin loiosxn vì không giúp được cả bài.
CHÚC BẠN HỌC GIỎI !!!
MÌNH TÌM RA CÁCH LÀM CÂU 3b RỒI !!!
5x+ 2x= 45+ 20: 15
5x+ 2x= 45+ \(\frac{4}{3}\)
5x+ 2x= \(\frac{139}{3}\)
(5+ 2)x=\(\frac{139}{3}\)
7x =\(\frac{139}{3}\)
x =\(\frac{139}{3}\): 7
x =\(\frac{139}{21}\)
CHÚC BẠN HỌC GIỎI !!!
a) \(\left(x-1\right)^2=49\)
\(\Rightarrow\left(x-1\right)^2=7^2=\left(-7\right)^2\)
\(\Rightarrow x-1=7\) hoặc \(x-1=-7\)
\(x=7+1=8\) \(x=-7+1=-6\)
Vậy x = 8 hoặc x = - 6
b) \(3\cdot\left(13-x\right)^2=27\)
\(\left(13-x\right)^2=27\div3=9\)
\(\Rightarrow\left(13-x\right)^2=3^2=\left(-3\right)^2\)
\(\Rightarrow13-x=3\) hoặc \(13-x=-3\)
\(x=13-3=10\) \(x=13+3=16\)
Vậy x = 10 hoặc x = 16
c) \(164-\left(15-x\right)^3=100\)
\(\left(15-x\right)^3=164-100=64\)
\(\Rightarrow\left(15-x\right)^3=4^3\)
\(\Rightarrow15-x=4\)
\(x=15-4=11\)
Vậy x = 11
d) \(\left(x+3\right)^3-15=210\)
\(\left(x+3\right)^3=210+15=225\)
\(\Rightarrow\left(x+3\right)^3=...\)
Tương tự mũ lẻ cậu nhé
e) \(x^2\div4=16\)
\(x^2=16\cdot4=64\)
\(\Rightarrow x^2=8^2=\left(-8\right)^2\)
Vậy x = 8 hoặc x = - 8
a/\(\left(x-1\right)^2\)=49
\(\left(x-1\right)^2\)=\(7^2\)
=>x-1=7
x=7+1
x=8
b/3.\(\left(13-x\right)^2\)=27
\(\left(13-x\right)^2\)=27:3
\(\left(13-x\right)^2\)=9
\(\left(13-x\right)^2\)=\(3^2\)
=>13-x=3
x=13-3
x=10
c/164-\(\left(15-x\right)^3\)=100
\(\left(15-x\right)^3\)=164-100
\(\left(15-x\right)^3\)=64
\(\left(15-x\right)^3\)=\(4^3\)
=>15-x=4
x=15-4
x=11
d/\(\left(x+3\right)^3\)-15=210
\(\left(x+3\right)^3\)=210+15
\(\left(x+3\right)^3\)=225
sai đề bài câu d hay sao ý bạn ạ
chỉ có \(\left(x+3\right)^2\)thì mới tính được
e/\(x^2\):4=16
\(x^2\)=16.4
\(x^2\)=64
\(x^2\)=\(8^2\)
=>x=8
a, (-0,2)2 \(\times\) 5 - \(\dfrac{2^{13}\times27^3}{4^6\times9^5}\)
= 0,04 \(\times\) 5 - \(\dfrac{2^{13}\times3^9}{2^{12}\times3^{10}}\)
= 0,2 - \(\dfrac{2}{3}\)
= \(\dfrac{2}{10}\) - \(\dfrac{2}{3}\)
= - \(\dfrac{7}{15}\)
b, \(\dfrac{5^6+2^2.25^3+2^3.125^2}{26.5^6}\)
= \(\dfrac{5^6+4.5^6+8.5^6}{26.5^6}\)
= \(\dfrac{5^6.\left(1+4+8\right)}{26.5^6}\)
= \(\dfrac{1}{2}\)
Bài 1
a) \(x=x^5\)
\(x^5-x=0\)
\(x\left(x^4-1\right)=0\)
\(x=0\) hoặc \(x^4-1=0\)
* \(x^4-1=0\)
\(x^4=1\)
\(x=1\)
Vậy x = 0; x = 1
b) \(x^4=x^2\)
\(x^4-x^2=0\)
\(x^2\left(x^2-1\right)=0\)
\(x^2=0\) hoặc \(x^2-1=0\)
*) \(x^2=0\)
\(x=0\)
*) \(x^2-1=0\)
\(x^2=1\)
\(x=1\)
Vậy \(x=0\); \(x=1\)
c) \(\left(x-1\right)^3=x-1\)
\(\left(x-1\right)^3-\left(x-1\right)=0\)
\(\left(x-1\right)\left[\left(x-1\right)^2-1\right]=0\)
\(x-1=0\) hoặc \(\left(x-1\right)^2-1=0\)
*) \(x-1=0\)
\(x=1\)
*) \(\left(x-1\right)^2-1=0\)
\(\left(x-1\right)^2=1\)
\(x-1=1\) hoặc \(x-1=-1\)
**) \(x-1=1\)
\(x=2\)
**) \(x-1=-1\)
\(x=0\)
Vậy \(x=0\); \(x=1\); \(x=2\)
a) \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)
\(\Leftrightarrow\left(x^2+6x+9\right)-\left(x^2+4x-32\right)-1=0\)
\(\Leftrightarrow2x=-40\)
\(\Rightarrow x=-20\)
b) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)
\(\Leftrightarrow x^3+27-x^3+4x=15\)
\(\Leftrightarrow4x=-12\)
\(\Rightarrow x=-3\)
c) \(\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)
\(\Leftrightarrow\left(x^2-4x+4\right)-\left(x^2+6x+9\right)-\left(4x+4\right)=5\)
\(\Leftrightarrow-14x=14\)
\(\Rightarrow x=-1\)
d) \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)
\(\Leftrightarrow4x^2-9-\left(x^2-2x+1\right)-\left(3x^2-15x\right)=-44\)
\(\Leftrightarrow17x=-34\)
\(\Rightarrow x=-2\)
e) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)
\(\Leftrightarrow24x=24\)
\(\Rightarrow x=1\)
a=197
b=157
c=121
d=16400
a)197
b)157
c)121
d)16400